Chapter 7 - Section 7.5 - Conditional Probability and Independence - Exercises - Page 507: 38

$\dfrac{1}{1296}$

As per the definition of in-dependency, we can write $P(A_1 \cap A_2 \cap A_{3} \cap A_{4})=P(A_1) P(A_2)P(A_{3})P(A_{4})$ Since, $P(A_{1})=P(A_{2})=P(A_{3})=P(A_{4})=\dfrac{1}{6}$ So, $P(A_1 \cap A_2 \cap..........\cap A_{11})=(\dfrac{1}{6}) \times (\dfrac{1}{6}) \times(\dfrac{1}{6})\times (\dfrac{1}{6})$ or, $=\dfrac{1}{6^{4}}$ or, $=\dfrac{1}{1296}$